Straightedge and Compass Constructions in Spherical Geometry

Transcription

1 350 MATHEMATICS MAGAZINE Straightedge and Compass Constructions in Spherical Geometry DANIEL J. HEATH Pacific Lutheran University Tacoma, WA Mathematicians love constructions. They begin with simple questions that even children can understand, but that often have unexpectedly difficult answers. A complete classification of constructible points, involving a tour through analytic geometry and abstract algebra, is included in many, perhaps most, abstract algebra textbooks. In the most famous math book of all time, Euclid began with five basic assumptions about geometry and built upon those assumptions [2]. The first three of Euclid s assumptions can be seen as instructions in the use of straightedge and compass, two tools employed extensively by Greek mathematicians. From these three assumptions, a game can be made. Given any set of points, we may construct the line determined by any two of them, or we may construct the circle with one of them as the center and another on the circle itself. We may add to the set of points any point arising as an intersection of two constructed lines, two constructed circles, or one constructed line and one constructed circle. Since we must start the game with something, we generally begin with two points, say A and B. The unit of measure being arbitrary in Euclidean geometry, we assume that the distance from A to B is 1. The goal of the game is to decide what, in general, is constructible. Of course, to simplify the game, we may choose specific things that we want to construct, such as a regular pentagon, a 20 angle, or the like, and try to answer those specific questions rather than the general question. The Greeks were adept at the straightedge and compass game. For example, they were able to show that segments with rational numbers and square roots of positive rational numbers as lengths were constructible. However, they were stymied by some other values, such as 3p 2. The straightedge and compass game saw no major breakthroughs for centuries, until Gauss proved in 1796 that a regular 17-gon is constructible by showing that cos 2 17 p = 1+p p p p 34 2 p p p 17, together with the knowledge that sums, differences, products, quotients, and square roots of constructible numbers are also constructible. A few years later he extended this result to show that a regular 257-gon and a regular gon are also constructible [3]. Gauss results imply that a regular n-gon can be constructed if n = 2 k 0 3 k 1 5 k 2 17 k k k 5, (1) where k 0 2{0, 1, 2,...}, and each of the other k i 2{0, 1}. (In fact, we could throw into that equation any other Fermat prime to the 0 or 1 power, though at this time none are known.) Math. Mag. 87 (2014) doi: /math.mag c Mathematical Association of America

2 VOL. 87, NO. 5, DECEMBER The coup de grâce to the (Euclidean) constructibility game was delivered by Wantzel, who was able to classify completely those numbers that are and are not constructible [11]. Essentially the idea was to use analytic geometry to turn the geometric problem into an algebraic problem, and to use the tools of abstract algebra in particular the field structure of the plane considered as the set of complex numbers to solve the algebraic problem. This result is discussed in detail in many places, including many introductory texts on abstract algebra (such as [9]). Spherical geometry The infamous fifth postulate of Euclid, however, hogged more attention than the others, in particular more than the three that give rise to the construction game. Essentially, this postulate says that, given a line and a point not on the line, there is exactly one line through the point and parallel to the given line. Centuries of mathematicians debated whether this postulate should, in fact, be a theorem rather than a postulate. Finally, in the early 19th century, two mathematicians, Bolyai and Lobachevsky, independently showed that the postulate was independent, and that making a different assumption leads to another geometry, currently called hyperbolic geometry [1, 7]. Years later Riemann noted that another assumption is possible, leading to still other geometries, currently called elliptic or projective geometry, and double elliptic or spherical geometry [8]. To our knowledge, the first person to investigate constructions in non-euclidean geometry was al-buzjani, a Persian mathematician, over 1000 years ago. He studied spherical geometry and trigonometry, and demonstrated numerous constructions on the sphere. A great introduction to his work can be found in [10]. In his beautifully written Master s thesis, Kincaid [5] showed that constructible segment lengths on the sphere are closed under the operations we will show, together with a cosine relation. This cosine relation is a spherical analog of the Pythagorean Theorem. If a segment length is constructible, then a cosine relation holds; but the converse does not necessarily hold, so this relation cannot be used to solve the spherical constructions game the way Wantzel solved the Euclidean version of the game. Let A and B represent points in a given geometry. We use notation AB to represent the distance from A to B, AB to represent the geodesic (line) determined by A and B (though we shall use the words geodesic and line interchangeably), and AB to represent the circle with center A and radius AB. We use the notation AB to denote the segment defined by A and B, and note that, in the spherical case, there are two such segments; we shall use this notation to refer to the shorter of the two, and AB to refer to the longer. Unlike Euclidean planes, two spheres are not necessarily congruent to one another; their radii must be congruent for the spheres themselves to be congruent. Hence, in a strict sense we could discuss the spherical construction game separately for each sphere of radius 0 < r < 1. In addition, and again unlike the Euclidean plane, we shall show that the choice of starting conditions affects the outcome of the game. However, we note that, since two points that are distance x apart on a sphere of radius 1 are equivalent to two points that are distance rx apart on a sphere of radius r, we shall assume without loss of generality that spherical geometry is the set of points S 2 ={(x, y, z) : x 2 + y 2 + z 2 = 1}. We note that distance between two points on the sphere is measured by arc length rather than chord length, and since we are restricting ourselves to the unit sphere, that is equivalent to the measure of the central angle subtended by those points. Equivalently, the distance is the inverse cosine of the dot product of the two points considered as vectors.

3 352 MATHEMATICS MAGAZINE For each point A on the sphere, there is a unique point whose coordinates are just the negative coordinates of A. We call this point the antipode of A and denote it A 0. We also note that for any two non-antipodal points A and B, there are two points C and C 0 that are each distance /2 from each of A and B. These points are called poles of A and B. Since C and C 0 are also poles for any points on AB, we sometimes refer to them as poles for the line. We note that, as in the Euclidean case, we may use analytic geometry to transform geometric information into algebraic information. In particular, the equation of a line (geodesic) determined by two (non-antipodal) points A(x 0, y 0, z 0 ) and B(x 1, y 1, z 1 ) in spherical geometry is (y 0 z 1 y 1 z 0 ) x + (z 0 x 1 z 1 x 0 ) y + (x 0 y 1 x 1 y 0 ) z = 0, (2) which is equivalent to ( EA EB) hx, y, zi = 0. The equation of the circle centered at A and containing point B (which may be A 0 ) is x 0 x + y 0 y + z 0 z = x 0 x 1 + y 0 y 1 + z 0 z 1, (3) which is equivalent to EA hx, y, zi = EA EB. In contrast with the Euclidean case, there is no natural field structure on the sphere; in fact, it doesn t even have a Lie group structure. (This means that there isn t a natural way to define something like addition for points on the sphere.) There is, however, a natural group structure on the circle, which we will use. Each point (x, y) on the circle can be written uniquely in the form (cos,sin ) for some 2 (, ]. For a group operation, we use angle addition (modulo 2 ). In other words, (cos 1, sin 1 ) + (cos 2, sin 2 ) = (cos( ), sin( )). We let {A, B} be the set of starting conditions for the construction game. The ultimate goal is to decide what points on the sphere are constructible using only a (spherical) straightedge and compass. If the distance between points A and B is /4, then the starting conditions are called standard starting conditions. As we will see below, several different (most) starting conditions imply the existence of two constructible points of distance /4, so it makes sense to start with something that is in general constructible anyway. Some special starting conditions EXAMPLE 1. Let AB =, that is, the starting conditions are two antipodal points. Then the set of constructible points on the sphere is exactly {A, B}. This is easy to verify. We note that since antipodal points do not determine a line, we cannot construct AB. We then note that AB ={B}, since the set of points of distance from A is just the antipode, B. Similarly, BA ={A}. Hence AB \ BA ={}, and no new points are constructible. We call a set of starting conditions of this form trivial of type 1. EXAMPLE 2. Let AB = /2. Then there is a point C 2 S 2 (a pole of AB) so that AB = AC = BC = /2, and the set of constructible points on the sphere is exactly {A, A 0, B, B 0, C, C 0 }.

4 VOL. 87, NO. 5, DECEMBER To see this, we note that AB \ BA ={C, C 0 }, AC \ CA ={B, B 0 }, and BC \ CB ={A, A 0 }. Now it is elementary to check that AB = AB 0 = A 0 B = A 0 B 0 = AC = AC 0 = A 0 C = A 0 C 0 = BC = BC 0 = B 0 C = B 0 C 0, and so on. Although there are 6 = 15 2 pairs of points that we might use to construct lines and circles, still there are only three distinct non-point constructions. We call a set of starting conditions of this form trivial of type 2. Note that the constructible set for starting conditions that are trivial of type 2 contains three disjoint sets of constructible points that are trivial of type 1. This shows that spherical constructions cannot, in general, be reverse engineered. In other words, just because we can construct point C given points A and B does not mean that we can construct point B if given points A and C. EXAMPLE 3. Let AB = 2 /3. Then there is a unique point C 2 S 2 so that the set of constructible points is exactly {A, B, C}. To make calculations easier, we take A = (1, 0, 0) and B = ( AB = (x, y, z) 2 S 2 : z = 0, AB = (x, y, z) 2 S 2 : x = 1 2, and 1/2, p 3/2, 0). Then BA = (x, y, z) 2 S 2 : 1 x + p 3 y + 0z = p Then C = AB \ BA = ( 1/2, 3/2, 0). We note that AB \ AB ={B, C} and AB \ BA ={A, C}. It only remains to show that the intersection of any lines or circles determined by A, B, and/or C is one or two of A, B, and/or C. This is elementary to verify. We call a set of starting conditions of this form trivial of type 3. Any starting conditions that are not trivial of type 1, 2, or 3 are said to be non-trivial. We note that, in light of Examples 1, 2 and 3, the starting conditions change the outcome. Thus, in the spherical version of the construction game, unlike the Euclidean version, there cannot be said to be a general set of starting conditions. However, in the next section we will see that, except for these three examples, constructible points are in fact dense on the sphere. Main results In this section, we prove some results in a traditional theorem-proof format. Each proof proceeds by construction, and should be replicable by undergraduate readers by finding the equations involved (using (2) for lines and (3) for circles), and then solving those equations using elimination of variables. THEOREM 1. Let A and B be given so that AB < 2 /3 and AB 6= /2. The perpendicular bisector of AB is constructible. Proof. Essentially, we use Euclid s proof of Proposition 10, noting exceptions along the way.

5 354 MATHEMATICS MAGAZINE Without loss of generality, let A = (1, 0, 0) and B = (x 0, y 0, 0). The conditions on AB give us 1/2 < x 0 6= 0. Then ( AB \ BA = x 0, x 0 x0 2, ± (1 x ) 0)(2x 0 + 1). y x 0 The condition 1/2 < x 0 means that the set contains two distinct points; call them C and D. The condition x 0 6= 0 means that they are not antipodes, so they determine a line. Next we consider AB \ CD ( ) 1 + x 0 1 x 0 = ±,, We call these two points M and M 0. We use an analytic argument to see that M bisects AB. Since cos(ab) = x 0, cos(am) = p (1 x 0 )/2. The latter, solved for x 0, is just x 0 = 1 2 cos 2 (AM). A standard trigonometric identity now shows that 2AM = AB. We note also that M 0 is the bisector of AB. We could use an analytic argument to show that AB? MM 0, too, but a geometric argument is simpler. If two triangles have each of their three sides pairwise congruent to one another, then those two triangles are congruent. (This theorem is known to hold in each of Euclidean, hyperbolic, and elliptic geometries [4].) Because AM = BM has been shown, PC = QC holds by construction, and because MC = MC holds by reflexivity, we have 1AMC = 1BMC, and thus \AMC = \BMC. Since, in addition, the measures of \AMC and \BMC sum to 180, this shows that \AMC must be a right angle, completing the proof. THEOREM 2. Let A and B be given so that AB 6= and AB 6= 2 /3. Then the poles for AB are constructible. C A N M B Figure 1 Constructing a pole Proof. Without loss of generality, we let A = (1, 0, 0) and B = (x 0, y 0, 0). We may in addition take y 0 > 0, so that we have y 0 = 1 x0. 2 We first show that we may assume AB < 2 /3. Suppose, instead, that 2 /3 < AB <, so 1 < x 0 < 1/2. Note that AB \ BA ={A, C}, where (using the fact that x0 2 + y2 0 = 1) C = x 0 2 y0 2, 2x 0y 0, 0. In addition, AC = cos 1 ( EA EC) =

6 VOL. 87, NO. 5, DECEMBER cos 1 (x0 2 y0 2). Since x y2 0 = 1, we have that x 0 2 y0 2 = 2x Since 1 < x 0 < 1/2, we have that 1/2 < x0 2 y0 2 < 1. Together, these imply that 0 < AC < 2 /3. Hence there is a constructible point C on AB whose distance is less than 2 /3 from A. Any points constructible from {A, C} are thus constructible from {A, B}. If AB = /2, then Example 2 already showed how to construct the poles for AB. Hence we assume that /2 6= AB < 2 /3. Now we apply Theorem 1 to AB, obtaining perpendicular bisector ` and midpoint M, with ` = (x, y, z) 2 S 2 : 2x 0 1 x x 0 p 2x0 + 1 x + 2x 0 (1 x 0 )(2x 0 + 1) 1 + x 0 y + 0 z = 0, {M, M 0 }= ± 1 + x 0, 2 1 x 0, 0. 2 Note that AM < /3, so we may apply Theorem 1 to AM, obtaining perpendicular bisector m and midpoint N. We do so by merely incrementing subscripts, setting x 1 = p (1 + x0 )/2, and obtaining m = (x, y, z) 2 S 2 : 2x 1 1 x x 1 p 2x1 + 1 x + 2x 1 (1 x 1 )(2x 1 + 1) 1 + x 1 y + 0 z = 0. The two lines ` and m intersect in {C, C 0 }={(0, 0, ±1)}, as in FIGURE 1. It is easy to see that these are the poles for AB, completing the proof. COROLLARY. Let distinct points P, Q, and R be given, with Q 6= P 0 and PR 6= PQ. Then the poles of PQ are constructible. Proof. First we note that we may assume R 2 PQ, since PR \ PQ is a pair of points in PQ that are distance PR from P. Then we have either {P, Q}, {P, R}, or {Q, R} satisfy the hypothesis of Theorem 2, so the poles of PQ are constructible, or PQ = QR = /2, in which case Example 2 demonstrates how to construct the poles. This completes the proof. Let {A, B} be given with 0 < AB <. Define a bijection between points of AB and the real numbers (, ], so that (A) = 0, (A 0 ) =, and (X) = AX if X 6= A, A 0 is on the same component of AB\{A, A 0 } as B, or (X) = AX otherwise. In other words, (X) is just the signed distance ±AX, where the sign is positive if X is in the same direction as B and negative otherwise. The set of numbers (, ] form a group under addition modulo 2. For points X, Y 2 AB, we define X = 1 ( (X)) and X + Y = 1 ( (X) + (Y )). In this sense, the set of points in AB form an additive group, which we shall denote G. This is the circle group, in which elements of the circle are added via standard angle addition, where the angle is as measured from point A in the (angular) direction of point B. THEOREM 3. The constructible numbers in AB form a subgroup of G. Proof. If the set of constructible numbers is as in Example 1 or 3, the result is straightforward, so we assume that the set of constructible numbers has at least five members. We show that if X, Y 2 AB are constructible, then X = 1 ( (X)) and

7 356 MATHEMATICS MAGAZINE X + Y = 1 ( (X) + (Y )) are constructible. The proof then follows from associativity of G and the fact that A acts as the identity element. Again for ease of computation we assume A = (1, 0, 0) and that B = (x 0, y 0, 0). If X = A 0, then X = 1 ( (A 0 )) (mod 2 ) = 1 ( ) = X, so X is its own inverse. Let X 6= A, A 0 be a constructible point on AB, and write X = (x 1, y 1, 0). Consider AB \ AX ={(x 1, y 1, 0), (x 1, y 1, 0)}. The first of these points is X. Notice that ((x 1, y 1, 0)) = (X); in other words, X = (x 1, y 1, 0) = 1 ( (X)). This shows that X is constructible. Let X = (x 1, y 1, 0) and Y = (x 2, y 2, 0) be constructible points on AB. The poles for AB are constructible by the Corollary. One of the poles, which we denote C, is (0, 0, 1). Then AX \ AC = {(x 1, 0, ±y 1 )}. We denote these points D 0 and D 1. Then we obtain ( CD 0 [ CD 1 ) \ YC = {(x 1 x 2, x 1 y 2, ±y 1 ), ( x 1 x 2, x 1 y 2, ±y 1 )}, which we denote E i for i = 0, 1, 2, 3. Lastly, we consider YE i \ AX, which is the set (x 1 x 2 + y 1 y 2, y 1 x 2 + x 1 y 2, 0), ( x 1 x 2 y 1 y 2, y 1 x 2 x 1 y 2, 0), (x 1 x 2 y 1 y 2, y 1 x 2 + x 1 y 2, 0), ( x 1 x 2 + y 1 y 2, y 1 x 2 x 1 y 2, 0). We call these points F i for i = 0, 1, 2, 3; part of their construction process can be seen in FIGURE 2. Recall that AX = cos 1 (x 1 ) and AY = cos 1 (x 2 ). This gives us AF i = cos 1 (±x 1 x 2 ± y 1 y 2 ) = cos 1 (± cos(ax) cos(ay) ± sin(ax) sin(ay)) = cos 1 (cos(±ax ± AY)) =±AX ± AY, and thus the F i are the points ±X ± Y, and in particular one of them is X + Y, which is therefore constructible. C A D 0 E 0 X F 0 Y F 1 Figure 2 Adding intervals

8 VOL. 87, NO. 5, DECEMBER THEOREM 4. Let AB be neither /2, 2 /3, nor ; that is, assume the starting conditions are non-trivial. Then there exists a constructible point M 2 AB with AM = /4. Proof. Following the proof of Theorem 2, we construct the set {A, A 0, C, C 0, D, D 0 } so that AC = /2, where C is on the same component of AB\{A, A 0 } as B, and so that {D, D 0 }= AC \ CA, that is, D and D 0 are the poles for AC. We may assume AB < /2; for if not, E = B C is constructible by Theorem 3, and AE < /2. We may also assume that B falls on AC; for if not, B is constructible by Theorem 3, and falls on AC. Now we may also assume that AB < /4, since if AB > /4then E = C B is constructible, where AE < /4, and if AB = /4, we are done. Let F = C B. Since 0 < AB < AC = /2, we have that 0 < AF < /2, so we can find the perpendicular bisector ` of BF by Theorem 1, and the points {M, M 0 }= ` \ BF are the midpoints of BF and BF. Assume that M 2 BF (and M 0 2 BF). We will show that AM = CM, that is, AM = /4. Note that BM = FM and AB = FC by construction. Hence, in the group structure, M = (M B) + (B A) and M = (C F) + (F M), which implies that M A = C M; that is, MA = CM. Since AC = /2, this implies that AM = /4. This completes the proof. Since a segment of length /4 is constructible under any non-trivial starting conditions, we shall call starting conditions {A, B} with AB = /4 standard starting conditions. THEOREM 5. Assume standard starting conditions. Then the set of constructible points contained in AB is dense in AB. Proof. Let >0 be given, and let X 2 AB. We will show that there is a constructible point on AB whose distance from X is less than. Without loss of generality, we assume that (X) >0. As we have already seen, AB can be bisected by C 0, so that AC 0 = /8. Recursively, AC i 1 can be bisected by C i such that AC i = /(2 3+i ), and thus (C i ) = /(2 3+i ). Choose i 2 N such that /2 3+i <. By the Archimedean Property, there is some n 2 N such that (n 1) /(2 3+i )< (X) <n /(2 3+i ). Let W = n C i, which is constructible because C i is, and because constructible points are closed under (group) addition. Then WX <, completing the proof. THEOREM 6. Assume standard starting conditions. Then the set of constructible points is dense in S 2. Proof. Let >0 be given, and choose a point X 2 S 2. If X 2 AB we are done by Theorem 5, so assume not. Let C be a pole for AB, and let {Y, Y 0 }= CX \ AB. Without loss of generality, we assume that AY > 0. Repeat the argument of Theorem 5 to find some constructible point D such that DY < /2. Let E be a pole for CD, and {Z, Z 0 }= EX \ CD. Again without loss of generality, we assume that Z is closer to X (and therefore that Z 0 is closer to X 0 ). Repeat the argument of Theorem 5 to find some constructible point F so that FZ < /2; see FIGURE 3. We claim that FX <. We have that FX apple FZ + ZX by the triangle inequality, which holds in spherical geometry [4]. In addition, using the spherical law of sines we have sin(zx) = sin(cx) sin(\zcx) and sin(dy) = sin(\dcy). However, \ZCX = \DCY, and 0 < sin(cx) < 1 implies that sin(zx) < sin(dy). Again, since ZX and

9 358 MATHEMATICS MAGAZINE Z F X C D Y E Figure 3 Constructible points are dense DY are between 0 and /2, we have ZX < DY. It follows that FX < /2 + /2 =, and the proof is complete. More results and even more questions Not only have we not solved the constructibility question for the sphere, we doubt whether the question is answerable in general. (In a forthcoming paper, we give a necessary condition for a point in S 2 to be constructible, allowing us to give examples of non-constructible points, but casting doubt on points that fulfill the condition.) However, we can use the results thus far to explore the question of constructions in the projective plane, which we envision as the set of points satisfying the equation x 2 + y 2 + z 2 = 1, together with the equivalence relation (x, y, z) ( x, y, z). We note that the greatest distance between two points in this geometry is /2, so Examples 1 and 3 cannot occur here. However, we do have the following, which is the projective plane s analog of Example 2. EXAMPLE 4. Let {A, B} be the set of starting conditions for the construction game in the projective plane, and assume that AB = /2. Then there is a unique point C such that AB = AC = BC = /2, and the constructible set is exactly {A, B, C}. We call starting conditions of this form trivial, and any other starting conditions non-trivial. We can follow the proofs of the theorems of the previous section to show the following. THEOREM 7. Let {A, B} be a set of non-trivial starting conditions in the projective plane. Then the set of constructible points is dense in the projective plane. Note, however, that the results we have found in spherical and projective geometries do not easily generalize to the hyperbolic case. Is it possible to show that different sets of starting conditions lead to different sets of constructible points? Is there a length, like /4 in the sphere, that is always constructible (i.e., regardless of starting conditions) in hyperbolic geometry? Is the set of constructible points dense in the hyperbolic plane? These seem to be great questions for an undergraduate research project. Kincaid notes at the end of [5] that his results have an analog in the hyperbolic plane. Although his hyperbolic cosine relation (the hyperbolic analog of the Pythagorean Theorem) is transcendental and thus not particularly useful given our current tools, it doesn t suffer from the same geometric limitations as its spherical analog. Thus we

10 VOL. 87, NO. 5, DECEMBER can imagine that someday our tools will become sophisticated enough to solve the hyperbolic constructibility question in general. We would like to close by noting that both [6] and [10] give examples of constructions of the vertices of Platonic solids on the sphere. In particular, the constructions of the dodecahedron and icosahedron allow us, given standard starting conditions, to construct segments of length 2 /3 and 2 /5 in any geodesic PQ in the sphere (by placing a vertex of the construction at a pole of PQ and then joining that vertex to the 3 or 5 closest vertices by lines that divide PQ into 3 or 5 congruent segments). This means that we know how to construct segments of length m /n, where m is an integer, 0 < m < n, and n = 2 k 0 3 k 1 5 k 2 with k 0 2{0, 1, 2,...} and k i 2{0, 1} for i > 0. This should hearken back to Equation (1). Hence we conjecture: CONJECTURE. Given standard starting conditions, segments of length 2 17, 2 257, are constructible in spherical and projective geometries. REFERENCES 1. F. W. Bolyai, Tentamen, J. Bolyai s work on non-euclidean geometry appears as an appendix. 2. Euclid, The Thirteen Books of the Elements, translated by Sir T. L. Heath, Dover, New York, C. F. Gauss, Disquisitiones Arithmeticae, translated by A. A. Clarke, Yale University Press, New Haven, CT, D. C. Kay, College Geometry: A Discovery Approach, Addison Wesley Longman, Boston, J. M. Kincaid, Constructible numbers in double elliptic geometry, Master s thesis, Emporia State University, I. Lénárt, Non-Euclidean Adventures on the Lénárt Sphere, Key Curriculum Press, Emeryville, CA, N. I. Lobachevsky, Geometrical Investigations on the Theory of Parallel Lines, translated by G. N. Halstead, Reprinted in Bonola, NonEuclidean Geometry, 1912; Dover reprint, B. Riemann, On the hypotheses which lie at the bases of geometry, inaugural lecture transcribed by D. R. Wilkins, translated by W. K. Clifford, Nature 8 (183 & 184) (1998) 14 17, 36, doi.org/ /008014a0 9. J. J. Rotman, A First Course in Abstract Algebra, Prentice-Hall, Upper Saddle River, NJ, R. Sarhangi, An introduction to medieval spherical geometry for artists and artisans, proceedings of the Bridges Conference, London, 2006, P. Wantzel, Recherches sur les moyens de reconnaître si un Problème de Géométrie peut se résoudre avec la règle et le compas, Journal de Mathématiques Pures et Appliquées 1(2) (1837) Summary We examine straightedge and compass constructions in spherical geometry. We show via examples that the starting conditions affect the set of constructible points. Although current tools do not allow for a complete solution, we take a tour through group theory and real analysis to show that, in general, the set of constructible points is dense on the sphere. However, we conclude with more questions than answers. DANIEL J. HEATH received his Ph.D. in mathematics from the University of California at Davis under the direction of Abigail Thompson. He is an associate professor and the chair of mathematics at Pacific Lutheran University. His mathematical interests range from algebra to mathematical origami to topology, and his nonmathematical interests range from bicycling to jazz to mushroom hunting.